Question

The apparatus shown in figure consists of four glass columns
connected by horizontal sections . The height of two central
connected B and C are 49 cm each. The two outer columns A
and D are open to the atmosphere .A and C are maintained at
a temperature of $95^{\circ }C$ while the columns B and D are
maintained at $5^{\circ }C$.The height of the liquid in A and D
measured from the base line are 52.8 am and 51 am
respectively.Determine the linear coefficient of thermal
expansion of the liquid

Answer 1

Density of a liquid varies with temperature as
${\rho }_1{}^{\circ }=\left(\frac{{\rho }_0{}^{\circ }}{1+\gamma t}\right)$
Here, $\gamma$ is the coefficient of volume expansion of
temperature
In the figure
$h_1=52.8am,h_2=51cmandh=49cm$
Now, pressure at B = pressure at C
$p_0+h_1{\rho }_{95^{\circ }}g-h{\rho }_{5^{\circ }}g=p_0+h_2{\rho }_{5^{\circ }}g-h\rho 95^{\circ }g$
$\Rightarrow {\rho }_{95^{\circ }}(h_1+h)={\rho }_{5^{\circ }}(h_2+h)0$
$\Rightarrow \frac{{\rho }_{95^{\circ }}}{{\rho }_{5^{\circ }}}=\frac{h_2+h}{h_1+h}\Rightarrow \frac{\frac{{\rho }_{0^{\circ }}}{1+95\gamma }}{\frac{{\rho }_{0^{\circ }}}{1+5\gamma }}=\frac{h_2+h}{h_1+h}$
$\Rightarrow \frac{1+5\gamma }{1+95\gamma }=\frac{51+49}{52.8+49}=\frac{100}{101.8}$
Solving this equation, we get
$\gamma =2\times 10^{-4}{/}^{\circ }$
$\therefore$ Coefficient of linear expansion of temperature,
$\alpha =\frac{\gamma }{3}6.7\times 10^{-5}{/}^{\circ }C$